Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $r = \dfrac{6k + 18}{k + 5} \div \dfrac{k^2 + k - 6}{k + 5} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{6k + 18}{k + 5} \times \dfrac{k + 5}{k^2 + k - 6} $ First factor the quadratic. $r = \dfrac{6k + 18}{k + 5} \times \dfrac{k + 5}{(k + 3)(k - 2)} $ Then factor out any other terms. $r = \dfrac{6(k + 3)}{k + 5} \times \dfrac{k + 5}{(k + 3)(k - 2)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ 6(k + 3) \times (k + 5) } { (k + 5) \times (k + 3)(k - 2) } $ $r = \dfrac{ 6(k + 3)(k + 5)}{ (k + 5)(k + 3)(k - 2)} $ Notice that $(k + 5)$ and $(k + 3)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ 6\cancel{(k + 3)}(k + 5)}{ (k + 5)\cancel{(k + 3)}(k - 2)} $ We are dividing by $k + 3$ , so $k + 3 \neq 0$ Therefore, $k \neq -3$ $r = \dfrac{ 6\cancel{(k + 3)}\cancel{(k + 5)}}{ \cancel{(k + 5)}\cancel{(k + 3)}(k - 2)} $ We are dividing by $k + 5$ , so $k + 5 \neq 0$ Therefore, $k \neq -5$ $r = \dfrac{6}{k - 2} ; \space k \neq -3 ; \space k \neq -5 $